Physics Chapter 1 Answers

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Chapter 1: Introduction to Physics

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Answers to Even-Numbered Conceptual Questions

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2.

The quantity T + d does not make sense physically, because it adds together variables that have physical dimensions. The quantity d/T does make sense, however; it could represent the distance by an object in the time T .

4.

(a) 10 s; (b) 10,000 s; (c) 1 s; (d) 10 s; (e) 10 s to 10 s.

7

17

8

9

Solutions to Problems and Conceptual Exercises 1.

Picture the Problem: This is simply a units conversion problem. Strategy: Multiply the given number by conversion factors to obtain the desired units. Solution: (a) Convert the units:

$114,000, 000 ×

(b) Convert the units again:

$114 $114,00 ,000,0 0,000 00 ×

1 gigadollars

gigadollars = 0.114 gigadollars

1× 109 dollars 1 teradollars 1 × 10 dollars 12

1.14 × 10−4 tera terado do = 1.14

Insight: The inside back cover of the textbook t extbook has a helpful chart of the t he metric prefixes.

2.

Picture the Problem: This is simply a units conversion problem. Strategy: Multiply the given number by conversion factors to obtain the desired units. Solution: (a) Convert the units:

(b) Convert the units again:

70 μ m × 70 μ m ×

1.0 × 10−6 m μ m

1.0 × 10−6 m μ m

= 7.0 × 10−5 m ×

1 km 1000 m

= 7.0 × 10−8 km

Master your semester with Scribd Free Foron 30this Days Sign up to vote title 3. Picture the Problem: This is simply a units conversion problem. Read & The New York Times Useful  Not useful the desired units. Strategy: Multiply the given number by conversion factors to obtain t extbook has a helpful chart of the t he metric prefixes. Insight: The inside back cover of the textbook

Cancel anytime.

Special offer for students: Only $4.99/month. Solution: Convert the units:

03

Gm 1× 109 m

×

3 × 108 m/s

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James S. Walker, Physics

Picture the Problem: This is a dimensional analysis question.

t he same manner as algebraic expressions. Strategy: Manipulate the dimensions in the

Sheet Music

Solution: 1. (a) Substitute dimensions f or or the variables:

2. (b) Substitute dimensions for the variables:

x = vt

⎛m⎞ ⎟ ( s ) = m ∴ The equation is dimensionally co ⎝s ⎠

m=⎜

x = 12 at

2

⎛m⎞ 2 s = m ∴ dimensionally consistent 2 ⎟( ) ⎝s ⎠

m = 12 ⎜ 3. (c) Substitute dimensions for the variables:

t =

2 x a

⇒ s=

m m s

2

= s 2 = s ∴ dimensionally co

Insight: The number 2 does not contribute any dimensions to the problem.

6.


t he same manner as algebraic expressions. Strategy: Manipulate the dimensions in the Solution: 1. (a) Substitute dimensions for the variables: 2. (b) Substitute dimensions for the variables:

3. (c) Substitute dimensions for the variables:

4. (d) Substitute dimensions for the variables:

⎛m⎞ ⎟ ( s ) = m Yes ⎝s⎠

vt = ⎜

⎛ m⎞ 2 ( s ) = m Yes 2 ⎟ ⎝s ⎠ m ⎛ m⎞ 2at = 2 ⎜ 2 ⎟ ( s ) = No s ⎝s ⎠ 1 2

at = 12 ⎜

v

2

2

a

=

( m s) m s2

2

=m

Yes

Insight: When squaring the velocity you must remember to square the dimensions of both the numerator (me the denominator (seconds).

Master your semester with Scribd : This is a dimensional analysis question. PictureYork the Problem & The7. New Times

Read Free Foron 30this Days Sign up to vote title



 Not useful Cancel anytime.

Useful

Special offer for Strategy: students: Only $4.99/month. Manipulate the dimensions in the t he same manner as algebraic expressions.

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t he same manner as algebraic expressions. Strategy: Manipulate the dimensions in the p v = 2ax

2

Solution: Substitute dimensions for the variables:

2

p ⎛m⎞ ⎛m ⎞ ⎜ s ⎟ = ⎜ s 2 ⎟( m ) ⎝ ⎠ ⎝ ⎠

m 2 = m p +1 therefore p = 1 Insight: The number 2 does not contribute any dimensions to the problem.

9.

Picture the Problem: This is a dimensional analysis question. Strategy: Manipulate the dimensions in the t he same manner as algebraic expressions. Solution: Substitute dimensions for the variables:

a = 2 x t

p

[L] [T]

2

= [L][T] p

[T]−2 = [T] p therefore p = −2 Insight: The number 2 does not contribute any dimensions to the problem.

10. Picture the Problem: This is a dimensional analysis question. t he same manner as algebraic expressions. Strategy: Manipulate the dimensions in the Solution: Substitute dimensions for the variables on both sides of the equation:

v = v0 + at

[L] [T] [L] [T]

= =

[L] [T] [L] [T]

+

[L] [T]2

[T]

It is dimensionally consisten

Insight: Two numbers must have the same dimensions in order to be added or subtracted.

11. Picture the Problem: This is a dimensional analysis question. Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Strategy: Manipulate the dimensions in the t he same manner as algebraic expressions. & The New York Times Not useful  Useful [L] Solution: Substitute dimensions for the variables,

Special offer for where students: $4.99/month. [M]Only represents the dimension of mass:

Cancel anytime.

F = ma = [M]

[T]2

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13. Picture the Problem: This is a significant figures question.

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Strategy: Follow the given rules regarding the calculation and display of significant figures. Solution: (a) Round to the 3rd digit:

3.14159265358979 ⇒

(b) Round to the 5th digit:

3.14159265358979 ⇒ 3.1416

(c) Round to the 7th digit:

3.14159265358979 ⇒ 3.141593

3.14

Insight: It is important not to round numbers off too early when solving a problem because excessive roundin cause your answer to significantly differ from the true answer.

14. Picture the Problem: This is a significant figures question. Strategy: Follow the given rules regarding the calculation and display of significant figures.

2.9979 × 108 m/s ⇒

Solution: Round to the 3rd digit:

3.00 × 108 m/

Insight: It is important not to round numbers off too early when solving a problem because excessive roundin cause your answer to significantly differ from the true answer.

15. Picture the Problem: The parking lot is a rectangle.

144.3 m

Strategy: The perimeter of the parking lot is the sum of the lengths of its four sides. Apply the rule for addition of numbers: numbers: the number of decimal places after addition equals the smallest number of decimal places in any of the individual terms.

47.66 m

47.66

144.3 m

Solution: 1. Add the numbers:

144.3 + 47.66 + 144.3 + 47.66 m = 383.92

2. Round to the smallest number of decimal places in any of the individual terms:

383.92 m

Insight: Even if you changed the problem to

⇒ 383.9 m

( 2 × 144.3 m ) + ( 2 × 47.66 m )

you’d still have to report 383.9

answer; the 2 is considered an exact number so it’s the 144.3 m that limits the number of significant digits.

16. Picture the Problem: The weights of the fish are added. Master your semester with Scribd  Free For 30this Days the of decimal places Strategy: Apply the rule for addition of numbers, which states that Read Signnumber up to vote on title after additio  the smallest number of decimal places in any of the individual terms. Useful & The New York Times   Not useful the numbers: 1. Add Special offer for Solution: students: Only $4.99/month.

2. Round to the smallest number of decimal

Cancel anytime.

2.35 + 12.1 + 12.13 lb = 26.58 lb

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18. Picture the Problem: This is a significant figures question.

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Strategy: Apply the rule for multiplication of numbers, which states that the number of significant figures aft multiplication equals the number of significant figures in the least accurately known quantity. 2

Solution: 1. (a) Calculate the area and round to four significant figures:

A = π r 2 = π (14.37 m ) = 648.729144 m 2 ⇒

2. (b) Calculate the area and round to two significant figures:

A = π r 2 = π ( 3.8 m ) = 45.3645979 m 2 ⇒ 45

2

Insight: The number π is considered exact so it will never limit the number of significant digits you report in

19. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) Convert to feet per second:

m ⎞ ⎛ 3.28 ft ⎞ ft ⎛ ⎜ 23 ⎟ ⎜ ⎟ = 75 s ⎠⎝ 1 m ⎠ s ⎝

2. (b) Convert to miles per hour:

mi ⎛ m ⎞ ⎛ 1 mi ⎞ ⎛ 3600 s ⎞ ⎜ 23 ⎟ ⎜ ⎟⎜ ⎟ = 51 h ⎝ s ⎠ ⎝ 1609 m ⎠ ⎝ 1 hr ⎠

Insight: Conversion factors are conceptually equal to one, even though numerically they often equal somethi than one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend easy-to-comprehend fashi

20. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units.

⎛ 3.28 ft ⎞ ⎟ = 2069 ft ⎝ 1m ⎠

Solution: 1. (a) Find the length in feet:

( 631 m ) ⎜

2. Find the width in feet:

( 707 yd ) ⎜

3. Find the volume in cubic feet:

V = LWH = ( 2069 ft ) ( 2121 ft ) ( 110 ft) = 4.83×

⎛ 3 ft ⎞ ⎟ = 2121 ft 1 yd ⎝ ⎠

⎛ 1m ⎞ Master your semester 4. (b) Convert to cubic meters:with Scribd = 1.37 × 10 m  1 0 ft ) ⎜ ( 4.83 ×10 ⎟ on Read Free Days Sign ⎝up to vote title 3.28 ftFor ⎠ 30this  & The New York Times Useful  Not useful  Insight: Conversion factors are conceptually equal to one, even though numerically they often equal somethi 3

8

3

7

Cancel anytime.

3

Special offer for than students: one. Only They$4.99/month. are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend easy-to-comprehend fashi

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22. Picture the Problem: This is a units conversion problem.

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Strategy: Convert the frequency of cesium-133 given on page 4 to units of microseconds per megacycle, the by the number of megacycles to find find the elapsed time. time. Solution: Convert to micro seconds per megacycle and multiply by 1.5 megacycles:

⎛ ⎞ ⎛ 1× 106 cycles ⎞ ⎛ 1 μ s ⎞ 1s ⎟×⎜ ⎜ ⎟⎜ ⎟ = 108.7827757 −6 ⎝ 9,192, 631, 770 cycles ⎠ ⎝ Mcycle ⎠ ⎝ 1× 10 s ⎠ 108.7827757

μ s

Mcycle

× 1.5 Mcycle = 160 μ s = 1.6

Insight: Only two significant figures remain in the answer because of the 1.5 Mcycle figure given in the prob statement. The metric prefix conversions are considered considered exact and have an unlimited number of significant fi most other conversion factors have a limited number of significant figures.

23. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: Convert feet to kilometers:

⎛ 1 mi ⎞ ⎛ 1.609 km ⎞ ⎟⎜ ⎟ = 0.9788 km ⎝ 5280 ft ⎠ ⎝ 1 mi ⎠

( 3212 ft ) ⎜


24. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: Convert seconds to weeks:

⎛ 1 msg ⎞ ⎛ 3600 s ⎞ ⎛ 24 h ⎞ ⎛ 7 d ⎞ 4 msg ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 9 × 10 wk ⎝ 7 s ⎠ ⎝ h ⎠ ⎝ d ⎠ ⎝ wk ⎠

Insight: In this problem there is only one significant figure associated with the phrase, “7 seconds.”

25. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Master your semester with Scribd Foron 30this Days Sign to vote title 1 m up ⎛ Read ⎞Free Solution: Convert feet to meters: = 32.9 m (108 ft ) ⎜ ⎟ & The New York Times 3.28Useful ft ⎠  Not useful ⎝ Cancel anytime.

 

Special offer for Insight: students: Only $4.99/month. Conversion factors are conceptually equal to one, even though numerically they often equal somethi than one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend easy-to-comprehend fashi

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27. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) The speed must be greater than 55 km/h because 1 mi/h = 1.609 km/h. 2. (b) Convert the miles to kilometers:

mi ⎞ ⎛ 1.609 km ⎞ km ⎛ ⎜ 55 ⎟⎜ ⎟ = 88 h ⎠⎝ mi h ⎝ ⎠


28. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: Convert m/s to miles per hour:

⎛ 8 m ⎞ ⎛ 1 mi ⎞ ⎛ 3600 s ⎞ 8 ⎜ 3.00 × 10 ⎟⎜ ⎟⎜ ⎟ = 6.71× 10 s ⎠ ⎝ 1609 m ⎠ ⎝ 1 h ⎠ ⎝


29. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: Convert to ft per second per second:

m ⎞ ⎛ 3.28 ft ⎞ ft ⎛ ⎜ 98.1 2 ⎟ ⎜ ⎟ = 322 2 s ⎠⎝ 1 m ⎠ s ⎝



Strategy: Multiply the known quantity by appropriate conversion factors to change the units. In this problem “jiffy” corresponds to the time in seconds that it takes light to travel one centimeter. Solution: 1. (a) : Determine the magnitude of a jiffy:

Master your semester with Scribd & The New York Times 2. (b) Convert minutes to jiffys: Special offer for students: Only $4.99/month.

1s s ⎛ ⎞⎛ 1 m ⎞ −11 ⎜ ⎟⎜ ⎟ = 3.3357 × 10 8 cm ⎝ 2.9979 × 10 m ⎠ ⎝ 100 cm ⎠ −11 1 jiffy = 3.3357 × 10 s Read Free Foron 30this Days Sign up to vote title  0 s ⎞ ⎛ 1Not jiffyuseful⎞ ⎛ 6Useful  (1 minute) ⎜ Cancel ⎟ ⎜ anytime. −11 ⎟ = 1.7987 × 1 ⎝ 1 min ⎠ ⎝ 3.3357 × 10 s ⎠

Insight: Conversion factors are conceptually equal to one, even though numerically they often equal somethi

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32. Picture the Problem: The volume of the oil is spread out into a slick that is one molecule thick.

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Strategy: The volume of the slick equals its area times its thickness. Use this fact to find the area. Solution: Calculate the area for the known volume and thickness:

A =

V h

=

1.0 m3 ⎛ 1 μ m ⎞ ⎜ ⎟= 2 0.50 μ m ⎝ 1× 10 −6 m ⎠

Insight: Two million square meters is about 772 square miles!


Strategy: Multiply the known quantity by appropriate conversion factors factors to change the units. Then use a rati the factor change in part (b). Solution: 1. (a) Convert square inches to square meters:

2. (b) Calculate a ratio to find the new area:

⎛ 1 m2 ⎞ A = ( 8.5 in × 11 in ) ⎜ = 2 ⎟ ⎝ 1550 in ⎠ Anew Aold

=

Anew =

LnewW new LoldWold 1 4

=

( 12 Lold ) ( 12 W old ) LoldW old

=

Aold

Insight: If you learn to use ratios you can often make calculations like these very easily. Always put the new in the numerator and the old quantity in the denominator to make the new quantity easier to calculate at the en

34. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. Convert m/s to ft/s:

m ⎞ ⎛ 3.28 ft ⎞ ⎛ ⎜ 20.0 ⎟ ⎜ ⎟ = 65.6 ft/s s ⎠⎝ m ⎠ ⎝

2. (b) Convert m/s to mi/h:

m ⎞ ⎛ 1 mi ⎞ ⎛ 3600 s ⎞ ⎛ ⎜ 20.0 ⎟ ⎜ ⎟⎜ ⎟ s ⎠ ⎝ 1609 m ⎠ ⎝ 1 h ⎠ ⎝

Insight: Conversion factors are conceptually equal to one, even though numerically they often equal somethi than one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend easy-to-comprehend fashi Picture the Problem: This is with 35. your a units conversion problem. Master semester Scribd Read Free Foron 30 units. Days Sign up vote title Strategy: Multiply the known quantity by appropriate conversion factors to to change thethis & The New York Times  Useful  Not useful m ⎞ ⎛ 3.28 ft ⎞ ⎛

Convert meters to feet: Special offer for Solution: students: Only $4.99/month.

Cancel anytime.

 

⎜ 9.81 2 ⎟ ⎜ ⎟ = 32.2 ft/s s ⎠⎝ 1 m ⎠ ⎝

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37. Picture the Problem: Suppose all milk is purchased by the gallon in plastic containers.

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Strategy: There are about 300 million people in the United States, and if each of these were to drink a half ga milk every week, that’s about 25 gallons per person per year. Each plastic container is estimated estimated to weigh ab ounce. Solution: 1. (a) Multiply the quantities to make an estimate:

( 300 ×10

2. (b) Multiply the gallons by the weight of the plastic:

(1×10

10

6

people ) ( 25 gal/y/person ) = 7.5 × 109 gal/y ≅ 1010

⎛ 1 lb ⎞ 8 9 ⎟ = 6.25× 10 lb/y ≅ 10 lb/y ⎝ 16 oz ⎠

gal/y ) (1 oz/gal) ⎜

Insight: About half a billion pounds of plastic! Concerted recycling can prevent much of these containers containers fro clogging up our landfills.

38. Picture the Problem: The Earth is roughly a sphere rotating about its axis. Strategy: Use the fact the Earth spins once about its axis every 24 hours to find the estimated quantities. Solution: 1. (a) Divide distance by time:

v=

d t

=

3000 mi 3h

= 1000 1000 mi/h mi/h ≅ 103 mi/h mi/h

2. (b) Multiply speed by 24 hours:

circumference = vt = ( 3000 mi/h) ( 24 h) = 24,000 ,000 m

3. (c) Circumference equals 2πr :

r =

cir circum cumferenc rencee 24,00 24,000 0 mi = = 3800 3800 mi = 103 mi mi 2π 2π

Insight: These estimates are “in the ballpark.” The speed of a point on the equator is 1038 mi/h, the circumfe the equator is 24,900 mi, and the equatorial radius of the Earth is 3963 mi.

Master your semester with Scribd & The New York Times


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Useful  Not useful Cancel anytime. the Problem 39.for Picture : The lottery winnings are represented either by quarters or paper dollars. Special offer students: Only $4.99/month. Strategy: There are about 5 quarters and about 30 dollar bills per ounce.

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40. Picture the Problem: This is a dimensional analysis question.

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Strategy: Manipulate the dimensions in the same manner as algebraic expressions. Solution: 1. (a) Substitute dimensions for the variables:


3. (c) Substitute dimensions for the variables: 4. (d) Substitute dimensions for the variables:

v = a t

m ⎛m⎞ The equat quatiion is dim dimensi ensio onall ally co = ⎜ 2 ⎟ (s) = ∴ The s ⎝s ⎠ s 2 v = 12 at

m

⎛m⎞ 2 dimensio nsiona nallly consi onsist steent ≠ 12 ⎜ 2 ⎟ ( s ) = m ∴ NOT dim s ⎝s ⎠

m

t =

a v

⇒ s≠

m s2 ms

=

1 s

∴ NOT dimensionally cons

v 2 = 2a x

m2 m2 ⎛m⎞ 2 m = = ⎜ s 2 ⎟ ( ) s 2 ∴ dimensionally consistent s2 ⎝ ⎠

Insight: The number 2 does not contribute any dimensions to the problem.

41. Picture the Problem: This is a dimensional analysis question. Strategy: Manipulate the dimensions in the same manner as algebraic expressions. Solution: 1. (a) Substitute dimensions for the variables:

2

x t = ( m ) ( s ) = m ⋅ s 2

v2


x x

3. (c) Substitute dimensions for the variables:

2

t v

4. (d) Substitute dimensions for the variables:

t

=

= =

m2 s2 m

m s2 ms s

=

m s2

2

No

Yes

Yes =

m s2

Yes

Insight: One of the equations to be discussed later is for calculating centripetal acceleration, where we’ll note

acentripetal = v 2 r has units of acceleration, as we verified in part (b).

Master your semester with Scribd 42. Picture the Problem: This is a units conversion problem. Read Free Foron 30this Days Sign up to vote title Strategy: Multiply the known quantity by appropriate conversion factors to change the units. & The New York Times  Useful  Not useful Special offer for Solution: students: Only $4.99/month. 1. (a) Convert nm to mm:

Cancel anytime.

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⎛ 1×10 1 0 −9 m ⎞ ⎛ 1 mm ⎞ −4 ( 675 nm ) ⎜ ⎟⎜ ⎟ = 6.75 × 10 −3 × 1 n m 1 1 10 0 m ⎠ ⎝ ⎠⎝

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44. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) Convert seconds to minutes:

⎛ 605 beats ⎞ ⎛ 60 s ⎞ 4 ⎜ ⎟⎜ ⎟ = 3.63 × 10 beats/min s ⎝ ⎠ ⎝ min ⎠

2. (b) Convert beats to cycles:

631, 77 770 cy cycles ⎞ ⎛ 1s ⎞ ⎛ 9,192, 63 ⎜ ⎟⎜ ⎟ = 15,194,433 s ⎝ 605 beats ⎠ ⎝ ⎠


45. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units.

⎛ 1m ⎞ above the water ⎟ = 3.05 m ab ⎝ 3.28 ft ⎠

Solution: 1. Convert ten feet to m:

(10.0 ft ) ⎜

2. Convert ten knots to m/s:

(10.0 knot ) ⎜

⎛ 1.0 n.mi/hr ⎞ ⎛ 1.852 km ⎞ ⎛ 1000 m ⎞ ⎛ 1 hr ⎞ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎝ knot ⎠ ⎝ n.mi ⎠ ⎝ km ⎠ ⎝ 3600 s ⎠

Insight: If we were to describe the flying parameters of the helicopter in SI units, we would say it is flying “3

46. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) The acceleration must be greater than 14 ft/s2 because there are about 3 ft per meter.

m ⎞ ⎛ 3.28 ft ⎞ ft ⎛ ⎜14 2 ⎟ ⎜ ⎟ = 46 2 s ⎝ s ⎠⎝ m ⎠

2. (b) Convert m/s2 to ft/s2:

2

2

2

3. (c) Convert m/s to km/h :

m ⎞ ⎛ 1 km ⎞ ⎛ 3600 s ⎞ ⎛ ⎜14 2 ⎟ ⎜ ⎟⎜ ⎟ = 1.8 × 10 s ⎠ ⎝ 1000 m ⎠ ⎝ h ⎠ ⎝


47. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution:semester 1. (a) Convert m/s to mi/h: Scribd Master your with & The New York Times 2. (b) Convert m/s to m/ms

Special offer for students: Only $4.99/month.

m ⎞ ⎛ 1 mi ⎞ ⎛ 3600 s ⎞ m ⎛ 310 ⎜140 ⎟ ⎜ ⎟⎜ ⎟= sFree 6For 09 m 1title h ⎠ h ⎝Read ⎠to ⎝ 1vote ⎠ this ⎝ Days 30 Sign up on

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Useful  mCancel 10−3Not s ⎞ useful m ⎞ ⎛ 1×anytime. 140 14 0 .14 × 5.0 m ⎜ ⎟ = 0.14 ⎜ ⎟ s ⎠ ⎝ 1 ms ⎠ ms ⎝

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49. Picture the Problem: The probe rotates many times per minute.

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Strategy: Find the time it takes the probe to travel 150 yards and then determine how many rotations occurre that time interval. interval. Convert units to figure figure out the distance moved per revolution. revolution. Solution: 1. (a) Find the time to travel 150 yards:

s ⎛ 1 s ⎞ ⎛ 30.5 cm ⎞ ⎛ 3 ft ⎞ × 150 yd = 443 ⎟ = 2.95 ⎜ ⎟⎜ ⎟⎜ yd ⎝ 31 cm ⎠ ⎝ ft ⎠ ⎝ yd ⎠

2. Find the number of rotations in that time:

⎛ 7 rev ⎞ ⎛ 1 min ⎞ 51.6 rev rev = 51 compl complete ete re ⎜ ⎟⎜ ⎟ × 443 s = 51.6 ⎝ min ⎠ ⎝ 60 s ⎠

3. (b) Convert min/rev to ft/rev:

⎛ 1 min ⎞ ⎛ 60 s ⎞ ⎛ 31 cm ⎞ ⎛ 1 ft ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 8.7 ft/rev ⎝ 7 rev ⎠ ⎝ 1 min ⎠ ⎝ s ⎠ ⎝ 30.5 cm ⎠


50. Picture the Problem: This is a dimensional analysis question. Strategy: Find p to make the length dimensions match and q to make the time dimensions match. p

⎛ [L] ⎞ [T]q impl implie iess p = 1 =⎜ ⎟ [T] 2 [T] [T] ⎝ ⎠ [L]

Solution: 1. Make the length dimensions match:

1

2. Now make the time units match:

[T]2

=

[T]q [T]1

or [T]−2 = [T]q [T]−1 impl

Insight: Sometimes you can determine whether you’ve made a mistake in your calculations simply by checki ensure the dimensions work out correctly on both sides of your equations.

51. Picture the Problem: This is a dimensional analysis question.

Strategy: Find q to make the time dimensions match and then p to make the distance distance dimensions match. match. Rec 2 have dimensions of meters and g dimensions of m/s . q

Solution: 1. Make the time dimensions match:

⎛ [ L] ⎞ q p [T] = [ L ] ⎜ 2 ⎟ = [ L ] ([ L] [T]−2 ) impli ⎝ [T] ⎠ p

− 12

2. Now make the distance units match: Scribd Master your semester with   & The New York Times  Useful  Not useful Insight: Sometimes you can determine whether you’ve made a mistake in your calculations simply by checki

⎛ [L] ⎞ [T] = [ L ] ⎜ implies p = 2 ⎟ [T] Read Free Foron 30this Days Sign ⎝up to ⎠vote title p

theOnly dimensions work out correctly on both sides of your equations. Special offer for ensure students: $4.99/month.

Cancel anytime.

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7. Substitute in the numbers:

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6. Solve the quadratic equation for v0 :



v0 =

vd ) −ΔvΔt ± Δ v 2 Δt 2 − 4 ( Δt ) ( Δvd

2Δt

mi ⎞ ⎛ ⎛ 1h ⎞ ΔvΔt = ⎜ +7.9 ⎟ ( −13 s ) ⎜ ⎟ = − 0.0285 mi h ⎠ ⎝ ⎝ 3600 s ⎠ ⎛ 1h ⎞ Δt = ( −13 s ) ⎜ ⎟ = − 0.00361 h, and d = ⎝ 3600 s ⎠ 2

8. Find v0 :

v0 =

− ( − 0.0285 mi ) ± ( − 0.0285 mi) − 4 ( − 0.0285 2 ( − 0.00361 h )

v0 = 43 mi/h mi/h , − 51 mi/h

Insight: This was a very complex problem, but it does illustrate that it is necessary to know how to convert u order to properly solve problems. The units must be consistent with each other in order for the math to succe

53. Picture the Problem: The snowy cricket chirps at a rate that is linearly dependent upon the temperature.

Strategy: Take note of the g iven mathematical relationship relationship between the number of chirps N in 13 seconds and temperature T in Fahrenheit. Use the relationship to to determine the appropriate appropriate graph of N vs. T .

Solution: The given formula, N = T − 40, is a linear equation of the form y = mx + b. By comparing comparing the two expressions we see that N is akin to y, T is akin to x, the slope m = 1.00 chirps °F−1, and b = −40 °F. In the N vs. T , only two of the plots are linear, graphs of N linear, plots A and C, so we consider consider only those. Of those two, on an intercept of −40 °F, so we conclude that the correct plot is plot C. Insight: Plot B would be an appropriate depiction of a formula like N = T 2 − 40.

54. Picture the Problem: The snowy cricket chirps at a rate that is linearly dependent upon the temperature. Strategy: Use the given formula to determine the number of chirps N in 13 seconds, and then use that rate to time elapsed for the snowy cricket to chirp 12 times. Solution: 1. Find the number of chirps per second:

2. Find the time elapsed for 12 chirps:

N t

=

T − 40

13 s

1s 0.23 chirp

=

43 − 4 0 13 s

=

0.23 c s

× 12 chirps = 52 s

Master your semester with Insight: Note that we can employ eitherScribd the ratio 0.23 0.23 chirp chirp 1 s or the ratio 1 s 0.23 0.23 chirp chirp,, whichever  is mos Read Free Foron 30this Days Sign up to vote title  answering the particular question that is posed. & The New York Times  Useful  Not useful Cancel anytime.

Special offer students: Only $4.99/month. the Problem 55.for Picture : The snowy cricket chirps at a rate that is linearly dependent upon the temperature.

Strategy: Use the given formula to determine the temperature T that corresponds to the given number of chir

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56. Picture the Problem: The cesium atom oscillates many cycles during the time it takes the cricket to chirp on

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Strategy: Find the time in between chirps using the given formula and then find the number of cycles the undergoes during that time. Solution: 1. Find the time in between chirps:

2. Find the number of cesium atom cycles:

N t

=

T − 40.0

13.0 s

=

65.0 − 40.0 13.0 s

= 1.92

chirps s

⎞ 631, 77 770 cycles ⎞ ⎛ 1s ⎛ 9,192, 63 .78 × 109 cy ⎟ = 4.78 ⎜ ⎟⎜ s ⎝ ⎠ ⎝ 1.92 chirp ⎠

Insight: The number of significant figures might be limited by the precision of the numbers 13 and 40 that ar the description of the formula. In this case we interpreted them as exact and let the precision of the measurem limit the significant digits of our answer.



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Physics Chapter 1 Answers

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